Please help! No idea what this question is asking! Here are three pairs of half-
ID: 695155 • Letter: P
Question
Please help! No idea what this question is asking!
Here are three pairs of half-cells, each shown in standard reduction form: Standard reduction voltages are in the textbook. Write them after each half-cell. I. Zn2+(aq) +2e- Zn(s) 2. C03"(aq) + 1 e- Co2"(aq) 3.2 H+ + 2e- H2(g) SnA+(aq) + 2e- Sn2+(aq) A15+(aq) + 3 e- Al(s) Cu2+(aq) + 2e- Cu(s) (show a tube bubbling H2 gas onto a coiled Pt wire for this half-cell) For each pair, reverse the appropriate half-cell, and indicate the following in the diagrams (show ion flow with species and arrows on each side of the porous plug at the bottom of the cell): a) Voltage b) anode and cathode c) oxidation and reduction reactions d) (+) and (-) electrodes e) direction of electron flow f) number of electrons in balanced equation g) direction of ion flow (assume NO, and K ions as spectator ions if necessary) h) content of each half-cell i) substance used for electrodeExplanation / Answer
A):voltage can be calculated as
Voltage of cathode - voltage of anode
b):Anode is where oxidation occurs and cathode is where reduction occurs, therefore we have
1):here reduction of both Zn+2 and Sn+4 takes place but as Zn becomes neutral,it will act as anode as only oxidation is now possible aver there and Sn will act as cathode.
Same process can be applied to other two.
C): oxidation means removal of electron and reduction is addition of electron.Oxidation takes place at anode and reduction takes place at cathode.
Therefore reaction at anode is oxidation reaction and reaction at cathode is reduction reaction.
D):Anode is considered as negative terminal while as cathode is considered as positive terminal.
E}): as oxidation takes place at anode,which means that electrons are released at anode and thus they will flow to cathode in the external circuit but reverse will occur through salt bridge.and direction of current is opposite to that of electron flow.